341We can write the expression as
2|z - a| + 3 \left|z - \frac{4a}{3} \right| + 4\left|z - \frac{3a}{2} \right|
where
a = \frac{1+i}{2}
Notice that a , \frac{4a}{3} ,\frac{3a}{2} are collinear.
Now, you have to get clear that the candidates for minimizing the function will lie on this line. The reason, as Nishant sir pointed out, is that for any point not on the line, you can consider the foot of the perpendicular from that point to the line and this point will be closer to the points.
Now
LHS \ge 2\left(|z-a| + \left|z - \frac{4a}{3} \right|\right) + \left|z - \frac{4a}{3} \right| + 4\left|z - \frac{3a}{2} \right|
\ge 2\left| \frac{a}{3} \right| + \left|z - \frac{4a}{3} \right| + 4\left|z - \frac{3a}{2} \right|
(Since |x| + |y| ≥|x+y|)
The equality
|z - a| + \left|z - \frac{4a}{3} \right| \ge \left| \frac{a}{3} \right|
holds for those points lying between z=a and z=4a/3 (end points included). You can see that the expression gets minimised when z = 4a/3.
You can similarly experiment with any pair among these points and you will obtain the min at z=4a/3
which is 4 \left| \frac{a}{3} \right| = \frac{2\sqrt2}{3}
